D. World of Darkraft - 2
time limit per test 2 seconds
memory limit per test 256 megabytes
input standard input
output standard output
Roma found a new character in the game "World of Darkraft - 2". In this game the character fights monsters, finds the more and more advanced stuff that lets him fight stronger monsters.
The character can equip himself with k distinct types of items. Power of each item depends on its level (positive integer number). Initially the character has one 1-level item of each of the k types.
After the victory over the monster the character finds exactly one new randomly generated item. The generation process looks as follows. Firstly the type of the item is defined; each of the k types has the same probability. Then the level of the new item is defined. Let's assume that the level of player's item of the chosen type is equal to t at the moment. Level of the new item will be chosen uniformly among integers from segment [1; t + 1].
From the new item and the current player's item of the same type Roma chooses the best one (i.e. the one with greater level) and equips it (if both of them has the same level Roma choses any). The remaining item is sold for coins. Roma sells an item of level x of any type forx coins.
Help Roma determine the expected number of earned coins after the victory over n monsters.
Input
The first line contains two integers, n and k (1 ≤ n ≤ 105; 1 ≤ k ≤ 100).
Output
Print a real number — expected number of earned coins after victory over n monsters. The answer is considered correct if its relative or absolute error doesn't exceed 10 - 9.
Examples
input
1 3
output
1.0000000000
input
2 1
output
2.3333333333
input
10 2
output
15.9380768924
题解：
这道题真的是好题啊……
我们不难发现，每一种武器的金币贡献都是相互独立的，因此我们只需要处理一种武器的情况，最后*k即可
那么我们考虑转移，设dp数组为f[i][j]，表示带着等级为j的武器打i只怪，能得到的期望收益，
那么显然f[0][j]均为0，我们最终的目标就是求f[n][1]
接下来考虑转移，每一次打怪有1/k的概率掉落这种装备，有(k-1)/k概率不掉落，就无法从这种装备上得到收益
因此f[i][j]=（期望金币收益）/k+(k-1)*f[i-1][j]/k
而期望金币收益中，有1/(j+1)概率掉落高级装备，j/(j+1)概率掉落低级装备
因此 期望金币收益=(f[i-1][j+1]+j)/(j+1)+(f[i-1][j]+(j+1)/2)*j/(j+1)
其中(j+1)/2为低级装备的期望收益（平均数）
因此综上，f[i][j]=((f[i-1][j+1]+j)/(j+1.0)+j*(f[i-1][j]+(j+1.0)/2.0 )/(j+1.0))/(k*1.0)+(k-1)*f[i-1][j]/(k*1.0);
但是，这样是一个O(n2）的算法，时间超限，空间也可能超限，因此我们考虑优化
首先显然，f数组可以滚动，解决了空间问题
接着，我们发现1s内108=105*1000，因此我们第二次循环变成1000次以内就好了
那么，我们可不可以在n太大时把第二层n变成一个小数吗？这是正确的吗？
我们可以发现，如果j很大的话，装备升级的可能就很小
如果是f[i][i+1]，总的概率是（1/k）i*1/（n+1）！，而题目中说“The answer is considered correct if its relative or absolute error doesn't exceed 10 - 9.”
因此这样小的数据我们完全可以扔掉不要
事实上，我们只要枚举j从1到1000左右即可，这样复杂度就被降到了1s以内，这样我们就解决了本题
代码见下：
#include <cstdio> #include <cstring> #include <algorithm> #include <cmath> using namespace std; int n,k; double f[2][850]; int main() { scanf("%d%d",&n,&k); int t=min(n,800); int now=1; for(int i=1;i<=n;i++,now^=1) for(int j=t;j;j--) f[now][j]=((f[now^1][j+1]+j)/(j+1.0)+j*(f[now^1][j]+(j+1.0)/2.0 )/(j+1.0))/(k*1.0)+(k-1)*f[now^1][j]/(k*1.0); printf("%.10lf",k*f[now^1][1]); }